Question

If $\omega$ is a complex cube root of unity and $x=\omega -{\omega }^2-2$, find the value of $x^4+3x^3+2x^2-11x-6.$

Answer 1

${(x+2)}^2=(\omega -{\omega }^2)$

$x^2+4x+4={\omega }^2+{\omega }^4-2{\omega }^3={\omega }^2+\omega -2$

As ${\omega }^2+\omega +1=0$

${\omega }^2+\omega =-1$

${\omega }^3=1$

$x^2+4x+4=-1-2=-3$

$x^2+4x+7=0$

Now, by long division method, we can say that

$x^4+3x^3+2x^2-11x-6=(x^2+4x+7)(x^2-x-1)+1$

As $x=\omega -{\omega }^2-2$

$x^2+4x+7=0$

Therefore value of

$x^4+3x^3+2x^2-11x-6$ is $1$